Let's analyze each function to find its domain.

(a) $f(x) = \frac{1}{\sqrt{x} + |x|}$

For this function, the denominator $\sqrt{x} + |x|$ must be defined and non-zero. Here are the considerations:

1. Domain of the square root function $\sqrt{x}$: For the square root function to be real, $x \geq 0$.
2. Absolute value function $|x|$: Defined for all $x$, so no restrictions here.
3. Denominator non-zero: We need $\sqrt{x} + |x| \neq 0$.

Let's analyze $\sqrt{x} + |x|$:

• For $x \geq 0$, $|x| = x$, so the expression becomes $\sqrt{x} + x$. This is always greater than zero for $x \geq 0$, meaning the denominator is non-zero for $x \geq 0$.
• For $x < 0$, $|x| = -x$, but $\sqrt{x}$ is undefined for negative $x$.

Thus, the domain for this function is: $\boxed{x > 0}$ (Since $x = 0$ makes the denominator zero, it is excluded.)

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(b) $f(x) = \frac{1}{\sqrt{x^2 - x - 6}}$

For this function, we must ensure the denominator $\sqrt{x^2 - x - 6}$ is defined and non-zero. Therefore:

1. Square root domain: The expression inside the square root, $x^2 - x - 6$, must be non-negative, i.e., $x^2 - x - 6 \geq 0$.
2. Denominator non-zero: $\sqrt{x^2 - x - 6} \neq 0$.

Let's solve the inequality $x^2 - x - 6 \geq 0$:

• Factor the quadratic: $x^2 - x - 6 = (x - 3)(x + 2)$.
• Solve $(x - 3)(x + 2) \geq 0$.

To solve this inequality, determine the sign of the expression by testing intervals based on the roots $x = 3$ and $x = -2$:

• For $x < -2$, both factors are negative, so the product is positive.
• For $-2 < x < 3$, one factor is positive, the other is negative, so the product is negative.
• For $x > 3$, both factors are positive, so the product is positive.

Thus, $x^2 - x - 6 \geq 0$ for $x \in (-\infty, -2] \cup [3, \infty)$.

Now, exclude points where $x^2 - x - 6 = 0$ (i.e., $x = -2$ and $x = 3$), as they make the denominator zero.

Thus, the domain is: $\boxed{(-\infty, -2) \cup (3, \infty)}$

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Summary of Domains:

• (a) $f(x) = \frac{1}{\sqrt{x} + |x|}$: Domain is $x > 0$.
• (b) $f(x) = \frac{1}{\sqrt{x^2 - x - 6}}$: Domain is $(-\infty, -2) \cup (3, \infty)$.

Would you like a more detailed explanation of any part?

Related Questions:

1. What is the domain of $f(x) = \frac{1}{x + |x|}$?
2. How would the domain of $f(x) = \sqrt{x} + |x|$ differ?
3. What is the range of $f(x) = \frac{1}{\sqrt{x^2 - x - 6}}$?
4. How do you solve quadratic inequalities like $x^2 - x - 6 \geq 0$?
5. Can you explain how the absolute value function affects domains?

Tip:

Always check for values that make the denominator zero or lead to negative values inside square roots when finding domains.